AP EAMCET · PHYSICS · Oscillations
A particle executes simple harmonic motion with a time period \(0.6 \mathrm{~s}\) and amplitude \(10 \mathrm{~cm}\). Then the mean velocity of the particle over the time interval during which it travels a distance \(5 \mathrm{~cm}\) starting from the equilibrium position.
- A \(1 \mathrm{~ms}^{-1}\)
- B \(50 \mathrm{~cm} \mathrm{~s}^{-1}\)
- C \(10 \mathrm{~cm} \mathrm{~s}^{-1}\)
- D \(1 \mathrm{~cm} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(A) \(1 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
We have \[ \begin{aligned} & x=A \sin \omega t \Rightarrow 5=10 \sin \omega t \\ & \Rightarrow \frac{1}{2}=\sin \omega t \Rightarrow t=\frac{\pi}{6 \omega}=\frac{\pi \times T}{6 \times 2 \pi}=\frac{T}{12} \end{aligned} \] Then,…
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